Problem: You have found the following ages (in years) of 5 zebras. Those zebras were randomly selected from the 42 zebras at your local zoo: $ 8,\enspace 11,\enspace 17,\enspace 7,\enspace 19$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 42 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $5$ samples and divide by $5$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\overline{x}} = \dfrac{8 + 11 + 17 + 7 + 19}{{5}} = {12.4\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {19.36} + {1.96} + {21.16} + {29.16} + {43.56}} {{5 - 1}} $ {s^2} = \dfrac{{115.2}}{{4}} = {28.8\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{28.8\text{ years}^2}} = {5.4\text{ years}} $ We can estimate that the average zebra at the zoo is 12.4 years old. There is also a standard deviation of 5.4 years.